If α and β are the zeros of the polynomial f(x) = x2 + px + q,
Question: If $\alpha$ and $\beta$ are the zeros of the polynomial $f(x)=x^{2}+p x+q$, then a polynomial having $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is its zero is (a) $x^{2}+q x+p$ (b) $x^{2}-p x+q$ (c) $q x^{2}+p x+1$ (d) $p x^{2}+q x+1$ Solution: Let $\alpha$ and $\beta$ be the zeros of the polynomial $f(x)=x^{2}+p x+q$. Then, $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=-\frac{p}{1}$ $=-p$ And $\alpha \beta=\frac{\text { Constant term }}{\text { Coef...
Read More →If a + b + c = 9 and ab + bc + ca = 23,
Question: If $a+b+c=9$ and $a b+b c+c a=23$, find value of $a^{2}+b^{2}+c^{2}$ Solution: We know that, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$ $9^{2}=a^{2}+b^{2}+c^{2}+2(23)$ $81=a^{2}+b^{2}+c^{2}+46$ $a^{2}+b^{2}+c^{2}=81-46$ $a^{2}+b^{2}+c^{2}=35$ Hence, value of required expression $a^{2}+b^{2}+c^{2}=35$...
Read More →If a2 + b2 + c2
Question: If $a^{2}+b^{2}+c^{2}=16$ and $a b+b c+c a=10$, find the value of $a+b+c$ ? Solution: We know that, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$ $(x+y+z)^{2}=16+2(10)$ $(x+y+z)^{2}=36$ $(x+y+z)=\sqrt{36}$ $(x+y+z)=\pm 6$ Hence, value of required expression I; (a + b + c) = 8...
Read More →If the sum of the zeros of the polynomial f(x) = 2x3 − 3kx2 + 4x − 5 is 6,
Question: If the sum of the zeros of the polynomial $f(x)=2 x^{3}-3 k x^{2}+4 x-5$ is 6 , then the value of $k$ is (a) 2 (b) 4 (c) $-2$ (d) $-4$ Solution: Let $\alpha, \beta$ be the zeros of the polynomial $f(x)=2 x^{3}-3 k x^{2}+4 x-5$ and we are given that $\alpha+\beta+\gamma=6$ Then, $\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=-\frac{(-3 k)}{2}=\frac{3 k}{2}$ It is given that $\alpha+\beta+\gamma=6$ Substituting $\alpha+\beta+\gamma=\frac{3 k}{2...
Read More →If a + b + c = 0 and a2 + b2 + c2
Question: If $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=16$, find the value of $a b+b c+c a$ : Solution: We know that, $\left[\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right]$ $(0)^{2}=16+2(a b+b c+c a)$ $2(a b+b c+c a)=-16$ $a b+b c+c a=-8$ Hence, value of required express ab + bc + ca = - 8...
Read More →Use algebraic identities to expand the following algebraic equations.
Question: Use algebraic identities to expand the following algebraic equations. (i) $(a+b+c)^{2}+(a-b+c)^{2}$ (ii) $(a+b+c)^{2}-(a-b+c)^{2}$ (iii) $(a+b+c)^{2}+(a+b-c)^{2}+(a+b-c)^{2}$ (iv) $(2 x+p-c)^{2}-(2 x-p+c)^{2}$ (v) $\left(x^{2}+y^{2}+(-z)^{2}\right)-\left(x^{2}-y^{2}+z^{2}\right)^{2}$ Solution: (i) We have, $(a+b+c)^{2}+(a-b+c)^{2}=\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)+\left(a^{2}+(-b)^{2}+c^{2}-2 a b-2 b c+2 c a\right)$ $\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y ...
Read More →If one zero of the polynomial f(x) = (k2 + 4)x2 + 13x + 4k
Question: If one zero of the polynomial $f(x)=\left(k^{2}+4\right) x^{2}+13 x+4 k$ is reciprocal of the other, then $k=$ (a) 2 (b) $-2$ (c) 1 (d) $-1$ Solution: We are given $f(x)=\left(k^{2}+4\right) x^{2}+13 x+4 k$ then $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=\frac{-13}{k^{2}+4}$ $\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=\frac{4 k}{k^{2}+4}$ One root of the polynomial is reciprocal of the other. Then, we h...
Read More →If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7,
Question: If $\alpha, \beta$ are the zeros of the polynomial $p(x)=4 x^{2}+3 x+7$, then $\frac{1}{\alpha}+\frac{1}{\beta}$ is equal to (a) $\frac{7}{3}$ (b) $-\frac{7}{3}$ (c) $\frac{3}{7}$ (d) $-\frac{3}{7}$ Solution: Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $p(x)=4 x^{2}+3 x+7$ $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$ $=\frac{-3}{4}$ $\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=\frac{7}{4}$ ...
Read More →Write the following in the expand form:
Question: Write the following in the expand form: (i) $(a+2 b+c)^{2}$ (ii) $(2 a-3 b-c)^{2}$ (iii) $(-3 x+y+z)^{2}$ (iv) $(m+2 n-5 p)^{2}$ (v) $(2+x-2 y)^{2}$ (vi) $\left(a^{2}+b^{2}+c^{2}\right)^{2}$ (vii) $(a b+b c+c a)^{2}$ (viii) $(x / y+y / z+z / x)^{2}$ (ix) $(a / b c+b / a c+c / a b)^{2}$ (x) $(x+2 y+4 z)^{2}$ (xi) $(2 x-y+z)^{2}$ (xii) $(-2 x+3 y+2 z)^{2}$ Solution: (i) We have, $(a+2 b+c)^{2}=a^{2}+(2 b)^{2}+c^{2}+2 a(2 b)+2 a c+2(2 b) c$ $\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x...
Read More →If α, β are the zeros of the polynomial f
Question: If $\alpha, \beta$ are the zeros of the polynomial $f(x)=x^{2}+x+1$, then $\frac{1}{\alpha}+\frac{1}{\beta}=$ (a) 1 (b) $-1$ (c) 0 (d) None of these Solution: Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}+x+1$ $\alpha+\beta=-\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}$ $=\frac{-1}{1}=1$ $\alpha \times \beta=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $=\frac{1}{1}=1$ We have $=\frac{1}{\alpha}+\frac{1}{\beta}$...
Read More →If the graph of quadratic polynomial ax2 + bx + c cuts negative direction of y-axis,
Question: If the graph of quadratic polynomial $a x^{2}+b x+c$ cuts negative direction of $y$-axis, then what is the sign of $c$ ? Solution: Since graph of quadratic polynomial $f(x)=a x^{2}+b x+c$ cuts negative direction of $y$-axis So put $x=0$ to find the intersection point on $y$-axis $y=0+0+c=c$ So the point is $(0, c)$ Now it is given that the quadratic polynomial cuts negative direction of $y$ So $c0$...
Read More →If graph of quadratic polynomial ax2 + bx + c cuts positive direction of y-axis,
Question: If graph of quadratic polynomial $a x^{2}+b x+c$ cuts positive direction of $y$-axis, then what is the sign of $c$ ? Solution: If graph of quadratic polynomial $f(x)=a x^{2}+b x+c$ cuts positive direction of $y$-axis, then Put $x=0$ for the point of intersection of the polynomial and $y$-axis We have $y=0+0+c=c$ Since the point $(0, c)$ is above the $x$-axis Hence, the sign of $c$ is positive, that is $c0$...
Read More →Prove that a2 + b2 + c2 − ab − bc − ca
Question: Prove that $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is always non-negative for all values of $a$, $b$ and $c$. Solution: We have, $a^{2}+b^{2}+c^{2}-a b-b c-c a$ Multiply and divide by 2 $=2 / 2\left[a^{2}+b^{2}+c^{2}-a b-b c-c a\right]$ $=1 / 2\left[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a\right]$ $=1 / 2\left[a^{2}+a^{2}+b^{2}+b^{2}+c^{2}+c^{2}-2 a b-2 b c-2 c a\right]$ $=1 / 2\left[\left(a^{2}+b^{2}-2 a b\right)+\left(a^{2}+c^{2}-2 c a\right)+\left(b^{2}+c^{2}-2 b c\right)\right]$ $=1 / 2\l...
Read More →If f(x) is a polynomial such that f(a) f(b) < 0,
Question: If $f(x)$ is a polynomial such that $f(a) f(b)0$, then what is the number of zeros lying between $a$ and $b ?$ Solution: If $f(x)$ is a polynomial such that $f(a) f(b)0$ then this means the value of the polynomial are of different sign for $a$ to $b$ Hence, at least one zero will be lying between $a$ and $b$...
Read More →If a quadratic polynomial f(x) is not factorizable into linear factors,
Question: If a quadratic polynomialf(x) is not factorizable into linear factors, then it has no real zero. (True/False) Solution: When polynomial $f(x)=a x^{2}+b x+c$ is not factorizable then the curve $y=a x^{2}+b x+c$ does not touch $x$-axis. Parabola $y=a x^{2}+b x+c$ open upwards above the $x$ axis or open downwards below $x$-axis where $a0$ or $a0$ Hence, if quadratic polynomial $f(x)$ is not factorizable into linear factors then it has no real zeros. True....
Read More →If a quadratic polynomial f(x) is a square of a linear polynomial,
Question: If a quadratic polynomialf(x) is a square of a linear polynomial, then its two zeros are coincident. (True/False). Solution: The polynomial $f(x)=x^{2}=(x-0)(x-0)$ has two identical factors. The curve $y=x^{2}$ cuts $X$ axis at two coincident points that is exactly at one point. Hence, quadratic polynomial $f(x)$ is a square of linear polynomial then its two zeros are coincident. True...
Read More →If a quadratic polynomial f(x) is factorizable into linear distinct factors,
Question: If a quadratic polynomialf(x) is factorizable into linear distinct factors, then what is the total number of real and distinct zeros off(x)? Solution: If a quadratic polynomial $f(x)=a x^{2}+b x+c$ is factorized into linear polynomials then the total number of real and distinct zeros of $f(x)$ will be 2...
Read More →Simplify each of the following products:
Question: Simplify each of the following products: (i) $\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\frac{1}{4} a^{2}+9 b^{2}\right)$ (ii) $(m+n / 7)^{3}(m-n / 7)$ (iii) $(x / 2-2 / 5)(2 / 5-x / 2)-x^{2}+2 x$ (iv) $\left(x^{2}+x-2\right)\left(x^{2}-x+2\right)$ (v) $\left(x^{3}-3 x-x\right)\left(x^{2}-3 x+1\right)$ (vi) $\left(2 x^{4}-4 x^{2}+1\right)\left(2 x^{4}-4 x^{2}-1\right)$ Solution: (i) We have, $\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\fra...
Read More →For what value of k, is −2 a zero of the polynomial 3x2 + 4x + 2k?
Question: For what value of $k$, is $-2$ a zero of the polynomial $3 x^{2}+4 x+2 k$ ? Solution: We know that if $x=\alpha$ is zero polynomial then $x-\alpha$ is a factor of $f(x)$ Since $-2$ is a factor of $f(x)$. Therefore $x+2$ is a factor of $f(x)$ Now, we divide $f(x)=3 x^{2}+4 x+2 k$ by $g(x)=x+2$ to find the value of k Now, Remainder $=0$ $2 k+4=0$ $2 k=-4$ $k=\frac{-4}{2}$ $k=-2$ Hence, the value of $k$ is $-2$...
Read More →For what value of k, is −2 a zero of the polynomial 3x2 + 4x + 2k?
Question: For what value of $k$, is $-2$ a zero of the polynomial $3 x^{2}+4 x+2 k$ ? Solution: We know that if $x=\alpha$ is zero polynomial then $x-\alpha$ is a factor of $f(x)$ Since $-2$ is a factor of $f(x)$. Therefore $x+2$ is a factor of $f(x)$ Now, we divide $f(x)=3 x^{2}+4 x+2 k$ by $g(x)=x+2$ to find the value of $k$ Now, Remainder $=0$ $2 k+4=0$ $2 k=-4$ $k=\frac{-4}{2}$ $k=-2$ Hence, the value of $k$ is $-2$...
Read More →For what value of k, is −3 a zero of the polynomial x2 + 11x + k?
Question: For what value of $k$, is $-3$ a zero of the polynomial $x^{2}+11 x+k$ ? Solution: We know that if $x=\alpha$ is zeros polynomial, then $x-\alpha$ is a factor of $f(x)$ Since $-3$ is zero of $f(x)$. Therefore $x+3$ is a factor of $f(x)$ Now, we divide $f(x)=x^{2}+11 x+k$ by $g(x)=x+3$ to find the value of $k$. Now, Remainder $=0$ $k-24=0$ $k=24$ Hence, the value of $k$ is 24 ....
Read More →If 3x - 7y = 10 and xy = -1,
Question: If $3 x-7 y=10$ and $x y=-1$, find the value of $9 x^{2}+49 y^{2}$. Solution: We have, $(2-7 y)^{2}=(3 x)^{2}+(-7 y)^{2}-2 * 3 x * 7 y$ $\Rightarrow(3 x-7 y)^{2}=9 x^{2}+49 y^{2}-42 x y \quad[$ Since, $3 x-7 y=10$ and $x y=-1]$ $\Rightarrow(10)^{2}=9 x^{2}+49 y^{2}+42$ $\Rightarrow 100-42=9 x^{2}+49 y^{2}$ $\Rightarrow 9 x^{2}+49 y^{2}=58$...
Read More →For what value of k, is 3 a zero of the polynomial 2x2 + x + k?
Question: For what value of $k$, is 3 a zero of the polynomial $2 x^{2}+x+k ?$ Solution: We know that if $x=\alpha$ is zero polynomial, and then $x-\alpha$ is a factor of $f(x)$ Since 3 is zero of $f(x)$ Therefore $x-3$ is a factor of $f(x)$ Now, we divide $f(x)=2 x^{2}+x+k$ by $g(x)=x-3$ to find the value of $k$ Now, remainder $=0$ $k+21=0$ $k=-21$ Hence, the value of $k$ is $-21$...
Read More →If 9x2 + 25y2 = 181 and xy
Question: If $9 x^{2}+25 y^{2}=181$ and $x y=-6$, find the value of $3 x+5 y$. Solution: We have, $(3 x+5 y)^{2}=(3 x)^{2}+(5 y)^{2}+2 * 3 x * 5 y$ $\Rightarrow(3 x+5 y)^{2}=9 x^{2}+25 y^{2}+30 x y$ $=181+30(-6) \quad\left[\right.$ Since, $9 x^{2}+25 y^{2}=181$ and $\left.x y=-6\right]$ $\Rightarrow(3 x+5 y)^{2}=1$ $\Rightarrow(3 x+5 y)^{2}=(\pm 1)^{2}$ $\Rightarrow 3 x+5 y=\pm 1$...
Read More →If α, β are the zeros of the polynomial 2y2 + 7y + 5,
Question: If $\alpha, \beta$ are the zeros of the polynomial $2 y^{2}+7 y+5$, write the value of $\alpha+\beta+\alpha \beta$. Solution: Let $\alpha$ and $\beta$ are the zeros of the polynomial $f(x)=2 y^{2}+7 y+5$. Then The sum of the zeros $=\frac{-\text { Coefficient of } y}{\text { Coefficient of } y^{2}}=\frac{-7}{2}$ The product of the zeros $=\frac{\text { Constant term }}{\text { Co-efficient of } y^{2}}=\frac{5}{2}$ Then the value of $\alpha+\beta+\alpha \beta$ is $\alpha+\beta+\alpha \b...
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