Solveeit Logo

Question

Question: If one root of the equation \({{x}^{2}}-30x+p=0\) is square of the other, then p is equal to. a. O...

If one root of the equation x230x+p=0{{x}^{2}}-30x+p=0 is square of the other, then p is equal to.
a. Only 125
b. 125, -216
c. 125, 215
d. Only 216

Explanation

Solution

We have been given information that one root is square of other, so we can consider one root as α\alpha , so the other root will be α2{{\alpha }^{2}}. We will use the concept of the sum of roots and the product of roots to solve this question. For general equation, ax2+bx+c=0a{{x}^{2}}+bx+c=0, we get the sum of roots as ba-\dfrac{b}{a} and the product of the root as ca\dfrac{c}{a}.

Complete step-by-step solution:
In the question, we have been given a quadratic equation, x230x+p=0{{x}^{2}}-30x+p=0 which has a variable p whose value we are supposed to find out. Apart from this information, we have been given that one root of this equation is equal to the square of the other root.
So, let us consider root as α\alpha , so the other root will become α2{{\alpha }^{2}}.
Now, we know that for a general quadratic equation, ax2+bx+c=0a{{x}^{2}}+bx+c=0, the sum of the roots is equal to ba-\dfrac{b}{a} and the product of the root is equal to ca\dfrac{c}{a}. We will use this concept to solve our question.
So, in the equation x230x+p=0{{x}^{2}}-30x+p=0, we have a = 1, b = -30 and c = p.
Thus, we can write the sum of the roots α and α2\alpha \text{ and }{{\alpha }^{2}} as
α+α2=ba\alpha +{{\alpha }^{2}}=-\dfrac{b}{a}
On putting a = 1 and b = -30, we get,
α+α2=301 α+α2=30 \begin{aligned} & \alpha +{{\alpha }^{2}}=-\dfrac{-30}{1} \\\ & \alpha +{{\alpha }^{2}}=30 \\\ \end{aligned}
On taking 30 to the LHS, we get,
α+α230=0\alpha +{{\alpha }^{2}}-30=0
We will solve this equation by using the middle term split method, to get the values of α\alpha .
α2+6α5α30=0 α(α+6)5(α+6)=0 (α5)(α+6)=0 α=5,6 \begin{aligned} & {{\alpha }^{2}}+6\alpha -5\alpha -30=0 \\\ & \alpha \left( \alpha +6 \right)-5\left( \alpha +6 \right)=0 \\\ & \left( \alpha -5 \right)\left( \alpha +6 \right)=0 \\\ & \alpha =5,-6 \\\ \end{aligned}
So, we get the values of α\alpha as 5 and -6.
Now, we know that for the given quadratic equation, x230x+p=0{{x}^{2}}-30x+p=0, we have a = 1, b = -30 and c = p.
Thus, the product of the roots α and α2\alpha \text{ and }{{\alpha }^{2}} can be written as,
α×α2=ca\alpha \times {{\alpha }^{2}}=\dfrac{c}{a}
On putting a = 1 and c = p, we get,
α×α2=p1 α3=p \begin{aligned} & \alpha \times {{\alpha }^{2}}=\dfrac{p}{1} \\\ & {{\alpha }^{3}}=p \\\ \end{aligned}
Now, we will substitute the values of α\alpha as 5 and -6 in the above equation and get the values of p from it.
Taking α=5\alpha =5, we get,
p=(5)3=125p={{\left( 5 \right)}^{3}}=125
Taking α=6\alpha =-6, we get,
p=(6)3=216p={{\left( -6 \right)}^{3}}=-216
Therefore, we get the values of p as 125 and -216.
Hence, option (b) is the correct answer.

Note: In this question, we used the direct formula of the sum and product of the roots as ba-\dfrac{b}{a} and ca\dfrac{c}{a}. We can also solve this question by using the formula of the root as b±b24ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. So, we will find the roots α and α2\alpha \text{ and }{{\alpha }^{2}}, and then on equating them, we will get the value of p.
Also, the students must take care not to make any calculation mistakes.