Solve the given two equations and select the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For equation $I$: $x^{2}-16x+63=0$We need two numbers whose product is $63$ and sum is $16$. These are $9$ and $7$.$x^{2}-9x-7x+63=0$$x(x-9)-7(x-9

Vedclass · Accepted Answer

if $x \ge y$

Vedclass · Answer

(B) For equation $I$: $x^{2}-16x+63=0$We need two numbers whose product is $63$ and sum is $16$. These are $9$ and $7$.$x^{2}-9x-7x+63=0$$x(x-9)-7(x-9)=0$$(x-9)(x-7)=0$So,$x = 9$ or $x = 7$.For equation $II$: $y^{2}-2y-35=0$We need two numbers whose product is $-35$ and sum is $-2$. These are $-7$ and $5$.$y^{2}-7y+5y-35=0$$y(y-7)+5(y-7)=0$$(y-7)(y+5)=0$So,$y = 7$ or $y = -5$.Comparing the values:If $x=9$,then $x > y$ (since $y$ is $7$ or $-5$).If $x=7$,then $x \ge y$ (since $y$ is $7$ or $-5$).Combining these,we get $x \ge y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-16x+63=0$
$II.$ $y^{2}-2y-35=0$

Similar Questions

For what value of $k$ can the quadratic polynomial $3z^{2} + 5z + k$ be factored into a product of real linear factors?

The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:

$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is equal to

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the roots of the equation $cx^2 + bx + a = 0$ are

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt{625}$
$II.$ $y = \sqrt{676}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-16x+63=0$$II.$ $y^{2}-2y-35=0$